///// 1
Consider the _FUNCTION_ template
template <typename T>
int spaceOf ()
{
int bytes = sizeof T;
return bytes / 4
+ bytes % 4 > 0;
}
Since C++ templates requires more intelligence from the environment
'we' have to tell the compiler about the actualy type:
So now:
class C {};
typdef void (*ptrFun)(int);
int idx = spaceOf<C>();
int jdx = spaceOf<ptrFunc>();
Except how do i achieve the same feat here for this fuction template:
class D {
public:
template<typename T> // here we go again....
D( T& t, void(T::*f)() )
{}
};
class UseD {
void someSpecialFunc() {};
D *ptrD;
public:
UseD() {
ptrD = new (std::nothrow) D(*this, &UseD::someSpecialFunc);
// validate
}
};
////////////// 2
For starters this is a 'C' ism and I suspect (scanned the ISO C++
standard and couldn't find anything) both are legal (compiler didn't
complain)?
a. typedef _MYSTRUCT {} MYSTRUCT;
b. typedef MYSTRUCT {} MYSTRUCT;
b at first appeared 'troubling' to me when I first encounted this and -
admittidely - I thought the compiler would complain. 5 1734 ma******@pegasus.cc.ucf.edu wrote: ///// 1 Consider the _FUNCTION_ template
template <typename T> int spaceOf () { int bytes = sizeof T;
int bytes = sizeof(T);
return bytes / 4 + bytes % 4 > 0; } Since C++ templates requires more intelligence from the environment 'we' have to tell the compiler about the actualy type: So now: class C {}; typdef void (*ptrFun)(int);
int idx = spaceOf<C>(); int jdx = spaceOf<ptrFunc>();
Except how do i achieve the same feat
What feat is that?
here for this fuction template:
class D { public: template<typename T> // here we go again.... D( T& t, void(T::*f)() ) {}
Are you looking to make use of 'T' here? Or of 't'? Or of 'f'? What
seems to be the problem?
};
class UseD { void someSpecialFunc() {}; D *ptrD; public: UseD() { ptrD = new (std::nothrow) D(*this, &UseD::someSpecialFunc); // validate } };
Since I don't know what "feat" it is, I am not sure what to tell you. ////////////// 2 For starters this is a 'C' ism and I suspect (scanned the ISO C++ standard and couldn't find anything) both are legal (compiler didn't complain)?
Is there a question here somewhere?
a. typedef _MYSTRUCT {} MYSTRUCT; b. typedef MYSTRUCT {} MYSTRUCT;
Did you mean
typedef struct BLAH {} BLAHBLAH;
? Without the keyword 'struct', it's _illegal_.
b at first appeared 'troubling' to me when I first encounted this and - admittidely - I thought the compiler would complain.
I feel your confusion.
V
> Except how do i achieve the same feat here for this fuction template: class D { public: template<typename T> // here we go again.... D( T& t, void(T::*f)() ) {}
};
class UseD { void someSpecialFunc() {}; D *ptrD; public: UseD() { ptrD = new (std::nothrow) D(*this, &UseD::someSpecialFunc);
You can't. 14.8.1.4 in 1997 draft standard.
// validate }
};
////////////// 2 For starters this is a 'C' ism and I suspect (scanned the ISO C++ standard and couldn't find anything) both are legal (compiler didn't complain)?
a. typedef _MYSTRUCT {} MYSTRUCT; b. typedef MYSTRUCT {} MYSTRUCT;
I think you mean
a. typedef struct _MYSTRUCT {} MYSTRUCT;
b. typedef struct MYSTRUCT {} MYSTRUCT;
b at first appeared 'troubling' to me when I first encounted this and - admittidely - I thought the compiler would complain.
Struct names live in a different 'namespace', that's a relic from C.
Don't do that:
struct MYSTRUCT
{
};
is enough in C++.
Jonathan
Victor, how are you :)
|| Are you looking to make use of 'T' here? Or of 't'? Or of 'f'?
What seems to be the problem?
I have to explicitly initialize T. Couldn't quite figure out how to
get there 'in' this case. The template functoin spaceOf was trivial in
comparison.
|| ? Without the keyword 'struct', it's _illegal_.
The distiction between a and b lies in the underscore for the type
MYSTRUCT. I was perusing souce code and saw a mix of
typedef _ST1 {} ST1 ; // note the underscore for the type
typedef ST2 {} ST2; // not the same here.
I thought to myself WTH. It was - apparently - deliberate. One or
two instance and I'd assume 'typo' but given 12/13 structs with 6/7
with underscore and the rest withouth, I became curious. I just prefer
consistency but first I thought there was something I was missing
| I think you mean
| a. typedef struct _MYSTRUCT {} MYSTRUCT;
| b. typedef struct MYSTRUCT {} MYSTRUCT;
Yes, I'm so sorry... Crud I can't belive I missed the keyword. I was
so hung up on the underscore that I mistyped. ma******@pegasus.cc.ucf.edu wrote: Victor, how are you :)
Thanks, I am fine. How are you? Are you looking to make use of 'T' here? Or of 't'? Or of 'f'? What seems to be the problem?
I have to explicitly initialize T.
I don't understand this statement. In a templated constructor you
cannot use explicit template argument syntax, you _have_to_ rely on
the template argument deduction from the function argument:
#include <typeinfo>
#include <iostream>
struct Foo {
template<class T> Foo(T t) {
std::cout << "Foo<" << typeid(T).name() << ">(" << t << ")\n";
}
};
int main() {
Foo foo(42);
}
(compile and run it, you should see something like)
---------
Foo<int>(42)
---------
Couldn't quite figure out how to get there 'in' this case. The template functoin spaceOf was trivial in comparison.
I don't understand why you'd need to explicitly specify the template
argument when it can be easily deduced by the compiler.
V This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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